the graph shows triangles def and klm.
is def congruent to klm? justify your answer.
yes, because a rotation 90° clockwise around d followed by a translation right 5 units maps def onto klm.
yes, because a reflection across the y - axis followed by a rotation 180° around e maps def onto klm.
no, because $ \overline { d e } $ and $ \overline { k l } $ do not have the same length.
no, because $ \overline { e f } $ and $ \overline { l m } $ do not have the same length.

Question

Accepted Answer

# Explanation:
## Step1: Calculate the length of $ \overline{DE} $
Use the distance formula $ d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} $. For $ D(-2,3) $ and $ E(-6,5) $, $ x_1=-2,y_1 = 3,x_2=-6,y_2 = 5 $.
$$
\begin{align*}
DE&=\sqrt{(-6+2)^2+(5 - 3)^2}\
&=\sqrt{(-4)^2+2^2}\
&=\sqrt{16 + 4}\
&=\sqrt{20}=2\sqrt{5}
\end{align*}
$$
## Step2: Calculate the length of $ \overline{KL} $
For $ K(3,2) $ and $ L(5,6) $, $ x_1 = 3,y_1=2,x_2 = 5,y_2 = 6 $.
$$
\begin{align*}
KL&=\sqrt{(5 - 3)^2+(6 - 2)^2}\
&=\sqrt{2^2+4^2}\
&=\sqrt{4+16}\
&=\sqrt{20}=2\sqrt{5}
\end{align*}
$$
## Step3: Calculate the length of $ \overline{EF} $
For $ E(-6,5) $ and $ F(-8,8) $, $ x_1=-6,y_1 = 5,x_2=-8,y_2 = 8 $.
$$
\begin{align*}
EF&=\sqrt{(-8 + 6)^2+(8 - 5)^2}\
&=\sqrt{(-2)^2+3^2}\
&=\sqrt{4+9}\
&=\sqrt{13}
\end{align*}
$$
## Step4: Calculate the length of $ \overline{LM} $
For $ L(5,6) $ and $ M(9,8) $, $ x_1 = 5,y_1=6,x_2 = 9,y_2 = 8 $.
$$
\begin{align*}
LM&=\sqrt{(9 - 5)^2+(8 - 6)^2}\
&=\sqrt{4^2+2^2}\
&=\sqrt{16+4}\
&=\sqrt{20}=2\sqrt{5}
\end{align*}
$$
Since $ EF=\sqrt{13}$ and $ LM = 2\sqrt{5}\approx4.47$, $ EF
eq LM$.

# Answer:
No, because $ \overline{EF} $ and $ \overline{LM} $ do not have the same length.

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of \( \overline{DE} \)

Step2: Calculate the length of \( \overline{KL} \)

Step3: Calculate the length of \( \overline{EF} \)

Step4: Calculate the length of \( \overline{LM} \)

Answer: